📂Abstract Algebra

Definition and Proof of Kronecker's Theorem for Extension Bodies

Definition of Extension Field ¹

For a field $F$, if there exists $E$ such that $F \le E$, then $E$ is called the Extension Field of $F$.

Kronecker’s Theorem

Assuming $f(x) \in F [ x ]$ is not a constant, there exists an extension field $E$ of $F$ and $\alpha \in E$ such that $f ( \alpha ) = 0$.

Description

An example of an extension field is that $\mathbb{C}$ is an extension field of $\mathbb{R}$. Kronecker’s Theorem implies that even if a polynomial does not have roots in $F$ immediately, expanding the domain to $E$ enables the existence of roots. The statement that blindly expanding the field would magically yield roots, without even knowing what $F$ looks like, is very much a mathematical theorem in spirit.

Proof

Part 1. Since $f(x)$ is not a constant function, it is uniquely factorized into irreducible elements over $F$, let one of these irreducible elements be $p(x)$. Then, the principal ideal $\left< p(x) \right>$ is a maximal ideal of $F [ x ]$, and $F [ x ] / \left< p(x) \right>$ becomes a field.

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Part 2. Existence of the Extension Field $E$

Defining the mapping $\psi : F \to F [ x ] / \left< p(x) \right>$ as follows makes $\psi$ naturally a homomorphism. $$ \psi (a) := a + \left< p(x) \right> $$

If for some $a,b \in F$, $\psi (a) = \psi (b)$ then $$ a + \left< p(x) \right> = b + \left< p(x) \right> $$ it implies $(b-a) \in \left< p(x) \right>$, which means that $(b-a)$ is a constant multiple of $p(x)$. But since $a, b \in F$ initially, $(b-a) \in F$ and, for $(b-a)$ to be a constant multiple of $p(x)$, it must be that $(b-a) = 0$. Therefore $\psi$ is injective, and $\psi$ becomes an isomorphism that sends every element of $F$ to a subfield of $F [ x ] / \left< p(x) \right>$. Defining specifically $E := F [ x ] / \left< p(x) \right>$ then makes $E$ an extension field of $F$.

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Part 3. Existence of the Root $\alpha \in E$

If we set $\alpha : = x + \left< p(x) \right>$ then initially $\alpha$ $\in E$. Define the substitution function $\phi_{\alpha} : F [ x ] \to E$ for this $\alpha$. If we explicitly present the chosen irreducible element as $p(x) := a_{0} + a_{1} x + \cdots + a_{n} x^{n}$

$$ \begin{align*} p ( \alpha) =& \phi_{\alpha} ( p(x) ) \\ =& a_{0} + a_{1} ( x + \left< p(x) \right> ) + \cdots + a_{n} ( x + \left< p(x) \right> )^n \\ =& a_{0} + a_{1} ( x + \left< p(x) \right> ) + \cdots + a_{n} ( x^n + \left< p(x) \right> ) \\ =& \left( a_{0} + a_{1} x + \cdots + a_{n} x^n \right) + \left< p(x) \right> \\ =& p(x) + \left< p(x) \right> \\ =& 0 + \left< p(x) \right> \end{align*} $$

Therefore, in $F [ x ] / \left< p(x) \right>$, $p( \alpha ) = 0$ and $\alpha$ satisfies $f ( \alpha ) = 0$.

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